1

Prof. Ji Chen

Notes 13 Transmission Lines

(Impedance Matching)

ECE 3317

Spring 2014

ZLZ0

ls

Z0s

d

2

Smith Chart

Impedance matching is very important to avoid reflected power, which causes a loss of efficiency and interference.

Zg

z

Sinusoidal source ZL

z = 0

Z0

S

L

0

0

LL

L

Z Z

Z Z

We will discuss two methods: Quarter-wave transformer

Single-stub matching

3

Quarter-Wave Transformer

Quarter-Wave Transformer: First consider a real load.

ZL = RLZ0 Z0T

Zin

/ 4l

0T0T

0T

tan

tanL

inL

Z jZ lZ Z

Z jZ l

2tan tan tan

2l l

20T

inL

ZZ

Z

20T

inL

ZZ

R realHence

4

20T

0L

ZZ

R

ZL = RLZ0 Z0T

Zin

/ 4l

Set 0inZ Z

This gives us 0T 0 LZ Z R

Hence

Quarter-Wave Transformer (cont.)

Example:

0

0

50

100

50 100 70.71

L

T

Z

Z

Z

5

Quarter-Wave Transformer (cont.)

Next, consider a general (complex) load impedance ZL.

Shunt (parallel) susceptance

s s LY jB jB Choose

YL = 1 / ZLZ0 Z0T

/ 4l

L L LY G jB

YLTOT = GLZ0 Z0T

/ 4l New model:

ZLTOT = 1 / GL (real)

Bs = -BL

6

Quarter-Wave Transformer (cont.)

Summary of quarter-wave transformer matching method

0T 0 / L

s L

Z Z G

B B

YL = GL + j BLZ0 Z0T

/ 4l

Ys = jBs

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Quarter-Wave Transformer (cont.)

Realization using a shorted stub

0 cots s s sB Y l

YL = GL + j BLZ0 Z0T

/ 4l

Bs = -BL

ls

Z0s

(An open-circuited stub could also be used.)

0 tans s s sX Z l

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Quarter-Wave Transformer with Extension

ZLZ0 Z0T

Zin(-d)

/ 4l

Z0

We choose the length d to make the input impedance Zin (-d) real.

We then use a quarter-wave transform to change the impedance to Z0.

d

9

Quarter-Wave Transformer with Extension (cont.)

ZLZ0 Z0T

/ 4l

Z0

d

Example

0 50[ ]Z

50 75 [ ]LZ j

1 (1.5)NLZ j

10

NLZ

NinZ d

0.176

0.250

0.250 0.176d

0.074d

0

4.3NinZ d SWR

Quarter-Wave Transformer with Extension (cont.)

Wavelengths towards generator

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4.3 50 4.3 215[ ]Nin inZ d Z d

0.074d

0 50 215TZ

0 103.7[ ]TZ

Z0T

/ 4l

Z0

d

0 50[ ]Z 50 75 [ ]LZ j

Quarter-Wave Transformer with Extension (cont.)

12

Single-Stub Matching

A susceptance is added at a distance d from the load.

ZL Y0 = 1 / Z0

d

s sY jB

1) We choose the distance d so that at this distance from the load

0in inY Y jB

2) We then choose the shunt susceptance so that

s inB B

(i.e., Gin = Y0)

inY

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Single-Stub Matching (cont.)

ZL Y0

d

s sY jB

0inY Y

The feeding transmission line on the left sees a perfect match.

s inB B

0in inY Y jB

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Realization using a shorted stub

ZL Z0

ls

Z0s

d

(An open-circuited stub could also be used.)

Single-Stub Matching (cont.)

15

ZL Z0

ls

Z0s

d

We use the Smith chart as an admittance calculator to determine the distance d.

1) Convert the load impedance to a load admittance YL.

2) Determine the distance d to make the normalized input conductance equal to 1.0.

3) Determine the required value of Bs to cancel Bin.

4) If desired, we can also use the Smith chart to find the stub length ls.

Single-Stub Matching (cont.)

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ZL Z0

ls

Z0s

d

Example0 50[ ]Z

100 100 [ ]LZ j

2 2NLZ j

10.25 0.25

2 2N

LY jj

/6 o0.62 0.62 30jL e 0

0

1

1

NL L

L NL L

Z Z Z

Z Z Z

Single-Stub Matching (cont.)

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2 2 NLZ j

X

X

0.178

Y 0.25 0. 5 2NL j

1 1.57j

1 1.57j

0.322

0.363

0.219

0.041

0.219 - 1.57

1.57 0.3

63

Ns

Ns

jY

Y

d

dj

at

at

Solution :

Add

or

0.170. 0.2198041

0.320. 0.3632041

Wavelengths toward loadWavelengths toward generator

Smith chart scale:

Use this one

Single-Stub Matching (cont.)

Re Rez z or

Im Imz z or

1inG 1NinG

X

X

18

Next, we find the length of the short-circuited stub: 1.57NsB

Rotate clockwise from S/C to desired Bs value.

S / C

0-j0.5

0-j1

0+j0.5

0+j1

0+j0

nY

Im z

Re z

0+j2

0-j2

0 1.57NsY j

Assume Z0s = Z0

Otherwise, we have to be careful with the normalization

(see the note below).

Admittance chart

Single-Stub Matching (cont.)

0 0

0 0

/

1.57 /

N Ns in s

s

B B Y Y

Y Y

Note: In general,

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0

0

1

tan

cot

cot

1.57 cot

1cot 1.57; tan 0.637

1.572

tan 0.637 0.567 [radians]

s s s

s s s

Ns s

s

s s

s s

Z jZ l

Y jY l

B l

l

l l

l l

Hence :

0.340 0.250sl

From the Smith chart:

0.0903sl

Analytically:

Single-Stub Matching (cont.)

0.09sl

S / C

X0 1.57j

0.09

O / C

Admittance chart

0.250

0.340

0

20

ZL

z0

UNMATCHED

Single-Stub Matching (cont.) 1+ 1.62L

1- 0.38L

z

1.621.55

0.0420.219

0.38

0.292

V z1.0

0.78

V 1

ZNL

X

1.62

0.78

0.219

0.042 (0.25 0.178 )

1.55Crank diagram

0.178

0.397

0

0.62L

z

21

ZL

z0

ZL

z

jBs

0.219 0

MATCHED

UNMATCHED

z

1.621.55

0.78

0.0420.219

SWR = 1.0

1+ 1.62L

1- 0.38L

z

1.621.55

0.0420.219

0.38

0.292

V z1.0

V z

0.78

Single-Stub Matching (cont.)

SWR = 4.26

V 1